Quote:
Originally Posted by Dominick Ferone
I must say I am confused by the spreadsheet, then again I am not to sure what I was looking at.
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This is a two-linkage-plus-glide configuration to launch projectiles using a motor. Launching with a motor is usually inefficient, as the load begins at rest and ends at maximum speed. A motor produces no power at rest or its maximum speed, but produces maximum power at approximately half speed. This setup allows a motor spinning at a constant speed to apply essentially constant power to the load as it accelerates from zero to top speed.
First, set up some sort of glide with a carriage to support the load. I was thinking of bearings on a rail topped by a light basket, but other configurations can be used. For purposes of the calculations on the spreadsheet, I set this rail along the y=0 axis. (which is the top of the frame in the layout diagram). In practice, you would probably rotate the entire assembly about forty-five degrees counter-clockwise.
I then placed the output shaft of the motor/gearbox at y=-11 units and x=7 units; these numbers are the first row in the spreadsheet and can be seen as the "hub" towards the lower right in the layout.
Attached to the motor/gearbox shaft is a linkage 7 units long; this is given in the second row, and is seen as the "star" around the hub in the layout.
Finally, connecting the motor linkage to the glide carriage is another linkage 20 units long. Ths length is specified in the third row. This is constrained to start at the end of the motor linkage and end along the y=0 axis. I calculate the left end point using the end of the motor linkage and the pythagorean theorem.
Spreadsheet, main data table:
Column A is the motor linkage rotation angle in degrees. I started at -10 so I could calculate acceleration at zero, and proceeded in five degree steps through 360. As I plotted this in the layout diagram, zero is when the motor linkage is horizontal to the left, positive angles rotate the motor linkage clockwise.
Column B is rotation angle in radians, used for sine and cosine calculations.
Columns C and D are the x and y coordinates of the motor/gearbox hub, repeated from row 1.
Columns E and F are the x and y coordinates of the end of the motor linkage. These are calculated from columns C and D plus the sin and cosine of the rotation angle times the motor arm length.
Columns G and H are the x and y coordinates of the glide end of the carriage linkage, calculated from the coordinates of the motor end, the constraint that y=0, and the pythagorean theorem.
Column I is the speed. Since I'm assuming constant rotation speed, this is simply the change in Xpos. I've taken the liberty of scaling up by 10 so that the graph looks nice.
Column J is the acceleration. Since I'm assuming constant rotation speed, this is simply the change in speed from the previous angle. Again, I scaled by 10.
Column K is the power, assuming a constant inertial load. Since power is force times speed, and force is mass times acceleration, this is simply speed times acceleration. Negative power implies that the load is being decelerated. Again, I scaled, this time by 1/5 to make a pretty graph.
The Position-Speed-Accel-Power plot simply plots columns G, I, J, and K vs column A.
The main power stroke is between about 37 degrees, where the load is stationary, and 170 degrees, where the load is at maximum speed. The carriage then stops by 250 degrees (resulting in the load flying off the end), and the portion from 250 degrees back to 37 degrees is the return stroke.
Note that the y location of the motor hub and the linkage lengths were selected by a manual tuning process to produce a flat power curve during the main stroke. x=7 was selected simply so hat the x output position starts at zero at the beginning of the power stroke after I had put in the other three numbers. You can change these numbers in rows 1, 2, and 3 to try different configurations, though you will need to turn off some of the axes limits on the graphs for them to be visible.
If you find graphs with sharp kinks in them, this is a sign that you are asking for something that is not going to fit, or require excessive forces. For example, set the carriage arm length to 18 units - you will get a large spike in acceleration and power at the very low angles. If you look at the layout diagram, you can see that this is because the carriage arm becomes vertical and must reverse direction very quickly. If you were to actually build this device, the carriage would probably overshoot this point or jam.
If you get lost, the original values were: -7, 11; 7; 20.
The secondary data table (columns M through AV) are tabulations of the x and y coordinates in a format more convenient to produce the layout diagram.